Equations of Motion of Systems with Internal Angular Momentum -II ManuelM. Dorado Dorado∗ CiRTA,ENITA, SL. S.A., Pol. Ind. Miguel Tres Yuste, Cantos12, Oeste, 28037 s/n Madrid, Spain Abstract Using the Euler’s equations and the Hamiltonian formulation, an attempt has been made to obtain the equations of motion of systems with internal angular momentum that are moving with respect to a reference frame when subjected to an interaction. This interaction involves the application of a torque that is permanently perpendicular to the internal angular momentum vector. mass) L , whose centre of mass moves at a constant velocity ν with respect to a reference frame which INTRODUCTION canbedefinedassoonastheinteractioninthe cylinder starts. It is aimed to obtain the equations The angular momentum of a many-particle sys- of motion to describe the dynamical behaviour of tem with respect to its centre of mass is known as the system from the instant that it undergoes an M the ”internal angular momentum” and is a prop- interaction, by applying a torque that is perma- erty of the system that is independent of the ob- nently perpendicular to the internal angular mo- server. Internal angular momentum is therefore mentum (See Fig.1). and attribute that characterizes a system in the When solving the problem, the following points same way as its mass or charge. In the case of a are taken into account: rigid body and particularly in the case of an ele- (a) The cylinder will continue spinning about its mentary particle, the internal angular momentum longitudinal axis at a constant angular veloc- is also referred to as ”spin”. ity ω throughout all the movement. In other To the author’s knowledge, the dynamical be- words, its internal angular momentum mod- haviour of systems with internal angular momen- ule is constant. The energy that the cylinder tum has not been systematically studied within the possesses as a result of its rotation about its framework of classical mechanics (See References longitudinal axis is consequently considered to [1]). This paper approaches this type of problem be internal and does not interfere with its dy- through a model comprising a rotating cylinder, namical behaviour. with constant velocity of rotation, around its longi- tudinal axis. The equations of motion are obtained (b) The derivative of the internal angular momen- using the Euler’s equations and by the Hamiltonian tum, L , with respect to a frame of axes of iner- procedure. tial reference (X, Y , Z) satisfies the equation Results coincide when the problem is solved us- ing vectorial algebra or Lagrangian formalism (un- dL dL × L published). However, as a result of changes in ”per- dt = dt + Ω (1) spective”, each method uncovers new peculiarities XY Z XY Z regarding the intimate nature of the system’s be- haviour. in which Ω is the rotation velocity of the frame linked to the solid (X, Y , Z)abouttheframe X Y Z FORMULATION OF THE PROBLEM of inertial reference axes ( , , ). (c) Only infinitesimal motions are considered that Let us consider a rigid cylindrical solid with internal are compatible with the system’s configura- angular momentum (with respect to its centre of tional variations brought about by the applied ∗ Email:E-mail: [email protected] [email protected] torque. (d) The total energy of the system is not explicitly (a) ω: the rotational velocity of the body around dependent on time. its longest axis (let us call it the intrinsic ro- tational velocity of the body). (e) No term is included that refers to the potential energy. According to hypothesis, the applied (b) Ω: the rotational velocity of the system of co- X Y torque is a null force acting on the system and ordinates associated with the body ( , , Z X Y the possibility of including a potential from ) viewed from the frame of inertia ( , , Z which the applied torque derives is unknown. ). We can write L and M , refered to (X, Y , Z), as follows: L = I1ω1e1 + I2ω2e2 + I3ω3e3 (3) Ω=Ω1e1 +Ω2e2 +Ω3e3 (4) M = M1e1 + M2e2 + M3e3 (5) So, the equation (2) takes the expression I1ω˙ 1 + I3ω3Ω2 − I2ω2Ω3 = M1 I2ω˙ 2 + I1ω1Ω3 − I3ω3Ω1 = M2 (6) I3ω˙ 3 + I2ω2Ω1 − I1ω1Ω2 = M3 Figure 1. Formulation of the problem These equations are known as modified Euler’s equations and can also be found in (1.e). HOW TO APPROACH THESE PROB- In the case treated above, we must substitute the LEMS THROUGH THE EULER’S EQUA- angular momentum TIONS? L = I3ω3e3 (7) The figure 1 shows a cylindrical rigid body with in (6): angular momentum, L , about its longest axis. We should consider two referential frames: one associ- I3ω3Ω2 = M1 ated with the solid (X, Y , Z), having the Z axis I3ω3Ω1 = M2 (8) on the direction of the body’s longest axis. We will I ω M refer this system of coordinates to a inertial frame 3 ˙ 3 = 3 (X, Y , Z). And then we get In a given instant, a torque M is applied on the M1 rigid body. Ω2 = We call Ω to the rotational velocity of the sys- I3ω3 tem of coordinates associated with the solid (X, M2 Ω1 = (9) Y , Z ), viewed from the inertial frame (X, Y , Z). I3ω3 From here, we will consider the system of coordi- M3 ω˙ 3 = nates (X , Y , Z ) all along the work. I3 Assuming this situation, it is known that If the torque is on the axis Y , M1 =0,M3 =0 dL dL and: = + Ω × L (2) dt dt 2 XY Z XY Z Ω =0 M2 Ω1 = (10) It is clear that we have two different velocities I3ω3 acting: ω˙ 3 =0 Where Ω1 coincides with the known velocity of Looking at the results arising from the previous precession. study, we can say that the body rotates following We have obtained that the body moves along a a circular orbit, with a radius given by the last circular trajectory with a rotational velocity Ω1, expression. This circular trajectory implies that but this is not enough to determinate the radius of the velocity vector must precess jointly with the the orbit. intrinsic angular momentum vector. From this last The principle of conservation for the energy conclusion it can be proved that the force needed should lead us to the expression for this radius. to cause the particle to draw a circular trajectory Assuming our initial hypothesis, the forces are is expressed as: applied perpendicular to the plane of movement of F mν × the solid and so, these forces do not produce any = Ω (15) work while the cylinder is moving. When the internal angular moment of the system We have two conditions that have to be satisfied does not coincide with one of the principal axes of by the body while rotating: inertia, the treatment of the problem is much more M complex, but as it will be seen in the next section, (a) Ω = L the results given above area completely general. (b) The total energy should remain constant. HAMILTONIAN FORMULATION. Before the torque is applied the energy of the EQUATIONS OF MOTION AND THEIR cylinder takes the following expression: SOLUTION 1 2 1 2 E = mνo + Iω0 (11) The equations of motion are obtained by the Hamil- 2 2 tonian formulation, where the independent vari- When the torque is acting on the cylinder about ables are the generalized coordinates and moments. q q t the line defined by the unit vector e2, it rotates To do this, a basis change of the frame ( ,˙, )to p q t along a trajectory, generally defined by r(t)and ( , , ) is made using the Legendre transformation. θ˙(t) that, in this case, coincides with Ω(t)andits From the function energy takes the following expression: H(p, q, t)= q˙ipi − L(q, q,˙ t) (16) i 1 2 2 2 1 2 E = m r˙ + r θ˙ + Iω0 (12) 2 2 asystemof2n + 1 equations is obtained. In this particular case our axes are principal axes ∂H ∂H δL q −p X M ˙i = ; ˙i = ; of inertia, therefore we have δt = 1,andthen, ∂tp ∂qi δ mr2θ˙ i δt ( )=0. (17) And therefore mr2θ˙ = const. ∂L ∂H − In the particular case in which θ˙ =Ωisacon- ∂t = ∂t stant, r has to be a constant (that is,r ˙ has to be If the last equation in (17) is excluded, a system zero) in order to satisfy this equation. We can con- n clude that the body moves in a circular orbit. The of 2 first order equations is obtained, known as energy takes the following expression: canonical Hamilton equations. By substituting H, the first order equations of 1 2 2 1 2 motion, are obtained. E = mr Ω + Iω0 (13) 2 2 It should be noted that the Hamiltonian formula- tion is developed for holonomous systems and forces Both expressions (11) and (13) have to represent derived from a potential that depends on the po- the same energy. Therefore, comparing the two ex- sition or from generalized potentials. A torque is pressions we can obtain the radius of the orbit: applied to the present system. The result of the ν0 forces is zero on the centre of mass and, therefore, r = (14) it is meaningless to refer to potential energy. Ω ∂V On the other hand, the Hamiltonian function The term − ∂r normally appears in this radial concept does remain meaningful. equation of motion and represents the force derived By using polar coordinate in the movement from a potential. In other words, plane, ˙ ∂V νr =˙r; vθ = rθ (18) mr mrθ˙2 − ¨ = ∂r (27) The Lagrangian is 1 2 2 2 In the present case, this term does not exist.